The sum of #n# terms of a geometric sequence can be written as:
\[s_n=t_1+t_2+t_3+\cdots+t_{n-1}+t_n=\sum_{k=1}^{n} t_k\]
If we want to calculate the sum for large values of #n#, it can be a lot of work to calculate this term by term and then add them. It is way quicker to use the general formula for the sum of the terms of a geometric sequence.
The sum of the first #n# terms of a geometric sequence #t_k# is equal to \[s_n =\sum_{k=1}^{n} t_k=t_1 \cdot \frac{r^n-1}{r-1}\]
We can prove this statement by using the fact that if we multiply a geometric sequence by the common ratio we end up with a new geometric sequence, that is equal to the first one apart from two terms.
A geometric sequence is of the form: #t_1#, #t_1 \cdot r#, #t_1 \cdot r^2#, #\ldots#.
Assume we want to add the first #n# terms of such a geometric sequence, we get the following sum:
\[s_n=t_1 + t_1 \cdot r + t_1\cdot r^2 + t_1 \cdot r^3 + \ldots + t_1 \cdot r^{n-2} + t_1 \cdot r^{n-1}\]
Now we multiply the sequence by#r#. We then get:
\[r \cdot s_n= t_1 \cdot r + t_1 \cdot r^2 + t_1 \cdot r^3 + \ldots + t_1 \cdot r^{n-1} + t_1 \cdot r^{n}\]
Now we subtract #s_n# from #r \cdot s_n#. This gives us:
\[\begin{array}{rcl}r \cdot s_n -s_n &=& (t_1 \cdot r + t_1\cdot r^2 + t_1 \cdot r^3 + \ldots + t_1 \cdot r^{n-2} t_1 \cdot r^{n-1} + t_1 \cdot r^{n}) \\ && - (t_1 + t_1 \cdot r + t_1\cdot r^2 + t_1 \cdot r^3 + \ldots + t_1 \cdot r^{n-2} + t_1 \cdot r^{n-1})\end{array}\]
If we reorder the terms, we get:
\[\begin{array}{rcl}(r-1) \cdot s_n &=& -t_1 + t_1 \cdot r - t_1 \cdot r+t_1 \cdot r^2 -t_1 \cdot r^2 + \ldots \\&& + t_1 \cdot r^{n-2} - t_1 \cdot r^{n-2} + t_1 \cdot r^{n-1} - t_1 \cdot r^{n-1} + t_1 \cdot r^n\end{array}\]
We see that most terms cancel each other out, we are left with:
\[(r-1) \cdot s_n = -t_1 + t_1 \cdot r^n\]
If we now move #(r-1)# to the other side, we have:
\[s_n = \frac{t_1 \cdot r^n-t_1}{(r-1)}=t_1 \cdot \frac{r^n-1}{r-1}\]
If #r# is less than #1#, it can be convenient to multiply the numerator and denominator by #-1# in the fraction.
The formula then becomes: \[s_n =\sum_{k=1}^{n} t_k=t_1 \cdot \frac{1-r^n}{1-r}\]
The advantage is that the denominator of the fraction is positive.
If #r=1#, then the fraction in the formula is not defined, since the denominator is equal to #0#. Then, the sequence is constant, hence, the sum of the first #n# terms is equal to #n\cdot t_1#.
It could happen that we want to know the sum of an infinite sequence. We call this an infinite series, which can be denoted by \[s_\infty = \sum_{k=1}^{\infty} t_k = \lim_{n\to\infty} \sum_{k=1}^{n} t_k\] For example, let #t_k=2^k# for #k\geq 1#. Then #t_1=2,\, t_2=4,\, t_3=8,\,\dots# The terms of this sequence get larger when #k# increases, which means the infinite series corresponding to this sequence is equal to infinity. In other words, \[ \sum_{k=1}^{\infty} 2^k=\infty\] Another example is #a_k=\left(\frac{1}{2}\right)^k# for #k\geq 1#. This gives #a_1=\frac{1}{2},\, a_2=\frac{1}{4},\, a_3=\frac{1}{8},\,\dots# The terms of this sequence get smaller when #k# increases, which means the infinite series corresponding to this sequence is a real number. In this case, \[ \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^k=1\]
Thus, to decide whether an infinite series is equal to infinity or to a real number, we need to know whether the terms of the sequence (in absolute value) are increasing or decreasing. If #r>1# or #r<-1# then the terms of the sequence are increasing in absolute value, so the infinite series is equal to plus or minus infinity. In this case, we say that the series is divergent.
On the other hand, if #-1<r<1#, then the terms of the sequence are decreasing in absolute value, so the infinite series is equal to a real number. In this case, we say that the series is convergent. To find the value of a convergent infinite series, we apply the limit from #n# to #\infty# to the formula for a geometric series. For a geometric sequence #t_k# with common ratio #r#, this gives \[s_{\infty}=\lim_{n\to\infty} \sum_{k=1}^{n} t_k = \lim_{n\to\infty} \left( t_1 \cdot \frac{r^n-1}{r-1}\right) = t_1 \cdot \frac{-1}{r-1}=\frac{-t_1}{r-1}\] since we know #\lim_{n\to\infty}r^n=0# because #-1<r<1#.
With this formula, we can easily calculate the sum of #n# terms of the geometric sequence if we know the number of terms, the first term and the common ratio.
From a geometric sequence #t# the initial term is #t_1=6# and the common ratio is #r=4#.
Calculate #\sum_{k=1}^{10} t_k#.
#\sum_{k=1}^{10} t_k=# #2097150#
The formula for the sum of the first #n# terms of a geometric sequence is equal to: \[\sum_{k=1}^n t_k=t_1 \cdot \frac{r^n-1}{r-1}\]
In this case we have
Which gives us:
\[\sum_{k=1}^{10} t_k=6 \cdot \frac{4^{10}-1}{4-1}=2097150\]